Let A be a commutative algebra of finite dimension, and if $A$ has no nilpotent elements other than $0$, is true that $A \cong \mathbb{C}^n$ ?
The question emerge to my mind, I thought that the finite dimension tell us that the scheme is Artinian (geometrically dimension 0).
I think the pattern is just a meeting of $n$ points but I have not managed to prove it.
Someone can enlighten me please ?
Thanks
On
It's true. If $A$ has no nilpotent elements, then it is semisimple. By the Artin-Wedderburn theorem, it must be a direct sum of matrix algebras over division rings. The only commutative such containing $\mathbb{C}$ is $\mathbb{C}$ itself.
This is probably overkill in that there is probably a more direct purely commutative proof.
On
A finitely generated commutative algebra with no nilpotents is always the ring of regular functions on some affine variety. Since your algebra is finite dimensional it is artinian and therefore has only finitely many maximal ideals, so since by the nullstellensatz maximal ideals are in 1-1 correspondence with points, it must be the ring of regular functions on some finite point space. The ring of regular functions on an $n$ point space is just $\mathbb{C}^n$
An artinian ring $A$ has only finitely many prime ideals, which are all maximal. Thus, by the Chinese remainder theorem, $$A / \mathfrak{n} \cong A / \mathfrak{m}_1 \times \cdots \times A / \mathfrak{m}_r$$ where $\mathfrak{m}_1, \ldots, \mathfrak{m}_r$ are the distinct prime/maximal ideals of $A$ and $\mathfrak{n}$ is the nilradical/Jacobson radical.
In particular, if the only nilpotent element of $A$ is $0$, then $A$ is a product of finitely many fields. Moreover, if $A$ is a finite $\mathbb{C}$-algebra, then each $A / \mathfrak{m}_i$ must also be a finite $\mathbb{C}$-algebra, hence, must be (isomorphic to) $\mathbb{C}$ itself.